Maths Assignment

Item 4B
Item 4B

Rachel Reiser
Maths C

Rachel Reiser
Maths C

Question 1 ab1+f'(x)2 dx

y = acosh(xa)

If: coshx=12ex+e-x Then: cosh(xa) = 12(exa+e-xa)

y = acosh(xa)
∴
y=a(exa+e-xa)2 y=a(exa+e-xa)2 dydx=f'x=ddxa(exa+e-xa)2

dydx=f'x=ddx12aexa+e-xa f'x=12a1aexa+-1ae-xa f'x=exa-e-xa2

f'x2=exa-e-xa22 f'x2=(12exa-12e-xa)(12exa-12e-xa) f'x2=14e2xa-14e0-14e0+14e-2xa f'x2=14e2xa-12+14e-2xa f'x2=14e2xa-2+e-2xa
Assuming the catenary is symmetrical, the entire length of the wire is equal to two multiplied by the length between x = b and x = a (or x = 0)
Entire arc length =2×0b1+f'(x)2 dx
Substitute the value for f'x2 into the formula above.
L =2×0b1+14e2xa-2+e-2xa dx
L =2×0b44+14e2xa-2+e-2xa dx
L =2×0b144+e2xa-2+e-2xa dx
L =2×0b142+e2xa+e-2xa dx
L=2×0b142+e2xa+e-2xa dx
L=2×140b2+e2xa+e-2xa dx
L=2×120b2+e2xa+e-2xa dx
Note: rule of perfect squares 2+e2xa+e-2xa=(exa+e-xa)2
L=0b(exa+e-xa)2 dx
L=0bexa+e-xa dx
L=aexa-ae-xa0b
L=aeba-ae-ba-(ae0a-ae-0a)
L=aeba-ae-ba-(a-a)
L=aeba-ae-ba
L=a(eba-e-ba)
∴ The length of the arc in terms of a and b is L=aeba-e-ba

Question 2 mdvdt=mg-kv2 Where: m=mass of body v=velocity g=acceleration due to gravity (9.8m/s) k=constant related to size and shape of object mdvdt=mg-kv2 mdv=mg-kv2 ×dt mdvmg-kv2=dt dt=mmg-kv2dv dt=m1mg-kv2dv mg-kv2 may also be expressed as mg2-kv22
=mg2-kv22
=mg2-kv2
∴ Difference of two squares: mg+kvmg-kv ∴dt=m1mg+kvmg-kvdv
RHS=m1mg+kvmg-kvdv
let 1mg+kvmg-kv=Amg+vk+Bmg-vk
∴1+0v=mg-vkA+mg+vkB
1+0v=mgA-vkA+mgB+vkB
1+0v=mgA+mgB+vkB-kA

∴1=mgA+mgB 0=kB-kA
1=mg(A+B) 0=k(B-A)
A+B=1mg --- 1 0=B-A --- 2 A=B --- 2
Sub 2 into 1:
2A=1mg
A=12mg
∴A=12mg B=12mg
1mg+kvmg-kv=Amg+vk+Bmg-vk
∴1mg+kvmg-kv=12mgmg+vk+12mgmg-vk
1mg+kvmg-kv=12mgmg+vk+12mgmg-vk
1mg+kvmg-kvdv=12mgmg+vkdv+12mgmg-vkdv

dt=m1mg+kvmg-kvdv

∴dt=m12mgmg+vk+12mgmg-vkdv